Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
pair<int, int> sdf;
long long arr[1000000];
int ar[100001];
bool preceed(int x, int y) { return x > y; }
int fact(int n) {
  if (n == 0 || n == 1)
    return 1;
  else
    return n * fact(n - 1);
}
int C(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); }
int inter(int a, int b, int c, int d) {
  if (max(a, c) > min(b, d))
    return 0;
  else {
    int i = min(b, d) - max(a, c) + 1;
    return i;
  }
}
int main() {
  long long n, s;
  cin >> n;
  if (n % 2 == 0 && n != 2)
    s = n / 2 + 1;
  else if (n == 2)
    s = 2;
  else {
    n;
    s = (n + 1) / 2;
  }
  cout << s << endl;
  long long a = 0, b = 1;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      a++;
    else
      b++;
    cout << a << " " << b << endl;
  }
  return 0;
}